Show line through point that is tangent to f(x)does not exist

In summary, the line through (2,7) that is tangent to the parabola y =x^2 +x does not have a slope that is equal to the general slope of the parabola.
  • #1
Painguy
120
0
1. show that there is no line through the point (2,7) that is tangent to the parabola y =x^2 +x



2. y-y1=m(x-x1)



3.
[itex]y'=f'(x)=2x+1[/itex]
[itex]m1=2x +1[/itex]
[itex]m1=2(2) +1 =5[/itex]

[itex]m2=((x^2 +x)-7)/(x-2)[/itex]

[itex]m1=m2?[/itex]
[itex]((x^2 +x)-7)/(x-2)=5[/itex]
[itex]x^2-4x+3=0[/itex]

[itex]2^2-4(2)+3=/=0[/itex]
[itex]-1=0[/itex]


I'm thinking that i would compare the slope of the line passing through (2,7) to the general slope of the parabola. The other thing i believe i could do is just plug 2 into y=x^2 +x. that would give me 6 which does not equal 7, but that does not involve using any of the material we're covering in class so that is out of the question.
 
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  • #2
One way to do it is to pick a random value for our hypothetical line to be tangent to y at: let's call it x1. (If that wasn't clear, we're assuming that we have a line that is tangent to the parabola at x-coordinate x1 and that also contains (2,7). We'll try to derive a contradiction to show that this is impossible). This is basically your idea to compare the slope of a random line through (2,7) to the general parabola slope.

Since our line is tangent to the graph of y at x1, then it has slope 2x1 + 1.
Since our line is tangent to the graph of y at x1, then it contains the point (x1, x1^2 + x1).
So we have a line that contains (x1,x1^2 + x1) and (2,7).
Calculating the slope of such a line using rise/over run, we get a new slope.
By assumption this new slope must equal the old slope, 2x1 + 1.
From there it is easy to set up a quadratic equation and show that it has no real solutions.
This shows that such a line cannot exist!

Your original idea to plug 2 into y does not work because we are only looking for a line that is tangent to y and contains (2,7). The only way your idea would work is if they were asking you to show that (2,7) is not a point on the parabola y.

I hope that is helpful!
 
  • #3
You are almost there. But the m1=m2 equation you want to try to solve is [itex]((x^2 +x)-7)/(x-2)=2x+1[/itex]. Don't substitute x=2 into m1. You don't know the tangent line hits the parabola at x=2. Draw a picture.
 

Related to Show line through point that is tangent to f(x)does not exist

1. What does it mean for a point to be tangent to a function?

When a point is tangent to a function, it means that the line passing through that point touches the curve of the function at only one point. This point of tangency is where the slope of the line is equal to the slope of the curve at that point.

2. Why might a point not have a tangent line to a function?

A point may not have a tangent line to a function if the slope of the curve at that point is undefined, or if the function is not continuous at that point. In other words, there is no unique line that can be drawn through the point and be tangent to the function.

3. Can a point be tangent to a function at more than one point?

No, a point can only be tangent to a function at one point. This is because the definition of a tangent line requires it to touch the function at only one point.

4. What is the significance of a point being tangent to a function?

When a point is tangent to a function, it tells us that the function is continuously differentiable at that point. This means that the function is smooth and has a well-defined slope at that point. It also helps us to understand the behavior of the function near that point.

5. How can we determine if a point is tangent to a function?

To determine if a point is tangent to a function, we can use the derivative of the function. If the derivative is equal to the slope of the line passing through the point, then the point is tangent to the function. Additionally, we can visually inspect the graph and see if the line passing through the point touches the curve at only one point.

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